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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p><dfn class="terminology">Solution</dfn> Consider</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_10.html">
\begin{equation}
({\bf A}-r{\bf I}) \vec{\xi}={\bf 0}.\tag{6.3.1}
\end{equation}
</div>
<p class="continuation">We require</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_10.html">
\begin{equation*}
|{\bf A}-r{\bf I} |=0 \to 
\left|
\begin{array}{ccc}
1-r &amp; 0 &amp; 0\\
2 &amp; 1-r &amp; -2\\
3 &amp; 2&amp; 1-r
\end{array}
\right|=0 \to
(1-r)(r^2-2r+2)=0 \to r_1=1-2 i, r_2=1+2i, r_3=1.
\end{equation*}
</div>
<p class="continuation">For <span class="process-math">\(r_1=1-2i\text{,}\)</span> from (<a href="" class="xref" data-knowl="./knowl/eq7_10.html" title="Equation 6.3.1">(6.3.1)</a>),</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_10.html">
\begin{equation*}
\left(
\begin{array}{ccc}
2 i &amp; 0 &amp; 0\\
2 &amp;2 i &amp; -2\\
3 &amp; 2 &amp; 2 i
\end{array}
\right) 
\left(
\begin{array}{c}
\xi^{(1)}_1\\
\xi^{(1)}_2\\
\xi^{(1)}_3
\end{array}
\right)={\bf 0} \to 
\begin{array}{c}
2 i \xi^{(1)}_1=0\\
2 \xi^{(1)}_1+2 i \xi^{(1)}_2-2 \xi^{(1)}_3=0\\
3\xi^{(1)}_1+2 \xi^{(1)}_2+2i\xi^{(1)}_3=0
\end{array}
\to
\xi^{(1)}_1=0,\quad \xi^{(1)}_3=i\xi^{(1)}_2.
\end{equation*}
</div>
<p class="continuation">Choose <span class="process-math">\(\xi^{(1)}_2=1\text{,}\)</span> the eigenvector is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_10.html">
\begin{equation*}
\xi^{(1)}=\left(\begin{array}{c}
0\\
1\\
i
\end{array}
\right)=\left(\begin{array}{c}
0\\
1\\
0
\end{array}
\right)+ i\left(\begin{array}{c}
0\\
0\\
1
\end{array}
\right)={\bf a}+i{\bf b}.
\end{equation*}
</div>
<p class="continuation">For <span class="process-math">\(r_2=1+2i\text{,}\)</span> <span class="process-math">\(\bar{\vec{\xi}}^{(1)}\)</span> is the eigenvector. Two linear independent solutions are</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_10.html">
\begin{equation*}
\begin{aligned}
&amp;{\bf x}^{(1)}=e^{\lambda t}({\bf a} \cos \mu t-{\bf b} \sin \mu t)=e^{t} 
\left[
\left(
\begin{array}{c}
0\\
1\\
0
\end{array}
\right) \cos 2 t-
\left(
\begin{array}{c}
0\\
0\\
1
\end{array}
\right) \sin 2t
\right],\\
&amp;{\bf x}^{(2)}=e^{\lambda t}({\bf b} \cos \mu t+{\bf a} \sin \mu t)=e^{t} 
\left[
\left(
\begin{array}{c}
0\\
0\\
1
\end{array}
\right) \cos 2 t+
\left(
\begin{array}{c}
0\\
1\\
0
\end{array}
\right) \sin 2t
\right].
\end{aligned}
\end{equation*}
</div>
<p class="continuation">For <span class="process-math">\(r_3=1\text{,}\)</span> the eigenvector is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_10.html">
\begin{equation*}
\vec{\xi}^{(3)}=\left(
\begin{array}{c}
2\\
-3\\
2
\end{array}
\right)
\end{equation*}
</div>
<p class="continuation">and the third solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_10.html">
\begin{equation*}
{\bf x}^{(3)}=\vec{\xi}^{(3)} e^{r_3 t}=e^t 
\left(
\begin{array}{c}
2\\
-3\\
2
\end{array}
\right).
\end{equation*}
</div>
<span class="incontext"><a href="sec6_3.html#p-269" class="internal">in-context</a></span>
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